6.02g Hooke's law: T = k*x or T = lambda*x/l

2 A uniform rod AB of length 0.5 m and mass 0.5 kg is freely hinged at A so that it can rotate in a vertical plane. Attached at B are two identical light elastic strings BC and BD each of natural length 0.5 m and stiffness \(2 \mathrm {~N} \mathrm {~m} ^ { - 1 }\). The ends C and D are fixed at the same horizontal level as A and with \(\mathrm { AC } = \mathrm { CD } = 0.5 \mathrm {~m}\). The system is shown in Fig. 2.1 with the angle \(\mathrm { BAC } = \theta\). You may assume that \(\frac { 1 } { 3 } \pi \leqslant \theta \leqslant \frac { 5 } { 3 } \pi\) so that both strings are taut. \begin{figure}[h] \end{figure}

1. Show that the length of BC in metres is \(\sin \frac { 1 } { 2 } \theta\).
2. Find the potential energy, \(V \mathrm {~J}\), of the system relative to AD in terms of \(\theta\). Hence show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 1.5 \sin \theta - 1.225 \cos \theta - \frac { 0.5 \sin \theta } { \sqrt { 1.25 - \cos \theta } } - 0.5 \cos \frac { 1 } { 2 } \theta .$$
3. Fig. 2.2 shows a graph of the function \(\mathrm { f } ( \theta ) = 1.5 \sin \theta - 1.225 \cos \theta - \frac { 0.5 \sin \theta } { \sqrt { 1.25 - \cos \theta } } - 0.5 \cos \frac { 1 } { 2 } \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-2_453_1264_2021_397} \captionsetup{labelformat=empty} \caption{Fig. 2.2}
   \end{figure} Use the graph both to estimate, correct to 1 decimal place, the values of \(\theta\) for which the system is in equilibrium and also to determine their stability.

4 A uniform lamina of mass \(m\) is in the shape of a sector of a circle of radius \(a\) and angle \(\frac { 1 } { 3 } \pi\). It can rotate freely in a vertical plane about a horizontal axis perpendicular to the lamina through its vertex O .

1. Show by integration that the moment of inertia of the lamina about the axis is \(\frac { 1 } { 2 } m a ^ { 2 }\).
2. State the distance of the centre of mass of the lamina from the axis. The lamina is released from rest when one of the straight edges is horizontal as shown in Fig. 4.1. After time \(t\), the line of symmetry of the lamina makes an angle \(\theta\) with the downward vertical. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-3_257_441_1475_322} \captionsetup{labelformat=empty} \caption{Fig. 4.1}
   \end{figure} \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-3_380_732_1635_1014} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
   \end{figure}
3. Show that \(\dot { \theta } ^ { 2 } = \frac { 4 g } { \pi a } ( 2 \cos \theta + 1 )\).
4. Find the greatest speed attained by any point on the lamina.
5. Find an expression for \(\ddot { \theta }\) in terms of \(\theta , a\) and \(g\). The lamina strikes a fixed peg at A where \(\mathrm { AO } = \frac { 3 } { 4 } a\) and is horizontal, as shown in Fig. 4.2. The collision reverses the direction of motion of the lamina and halves its angular speed.
6. Find the magnitude of the impulse that the peg gives to the lamina.
7. Determine the maximum value of \(\theta\) in the subsequent motion.

3 A uniform rigid rod AB of mass \(m\) and length \(2 a\) is freely hinged to a horizontal floor at A . The end B is attached to a light elastic string of modulus \(\lambda\) and natural length \(5 a\). The other end of the string is attached to a small, light, smooth ring C which can slide along a horizontal rail. The rail is a distance \(7 a\) above the floor and C is always vertically above B . The angle that AB makes with the floor is \(\theta\). The system is shown in Fig. 3. \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system and hence show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = a \cos \theta \left( m g - \frac { 4 \lambda } { 5 } ( 1 - \sin \theta ) \right) .$$
2. Show that there is a position of equilibrium when \(\theta = \frac { 1 } { 2 } \pi\) and determine whether or not it is stable. There are two further positions of equilibrium when \(0 < \theta < \pi\).
3. Find the magnitude of the tension in the string and the vertical force of the hinge on the rod in these positions.
4. Show that \(\lambda > \frac { 5 m g } { 4 }\).
5. Show that these positions of equilibrium are stable.

2 Fig. 2 shows a system in a vertical plane. A uniform rod AB of length \(2 a\) and mass \(m\) is freely hinged at A . The angle that AB makes with the horizontal is \(\theta\), where \(- \frac { 2 } { 3 } \pi < \theta < \frac { 2 } { 3 } \pi\). Attached at B is a light spring BC of natural length \(a\) and stiffness \(\frac { m g } { a }\). The other end of the spring is attached to a small light smooth ring C which can slide freely along a vertical rail. The rail is at a distance of \(a\) from A and the spring is always horizontal. \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system and hence show that \(\frac { \mathrm { d } V } { \mathrm {~d} \theta } = m g a \cos \theta ( 1 - 4 \sin \theta )\).
2. Hence find the positions of equilibrium of the system and investigate their stability.

3 Fig. 3 shows a uniform rigid rod AB of length \(2 a\) and mass \(2 m\). The rod is freely hinged at A so that it can rotate in a vertical plane. One end of a light inextensible string of length \(l\) is attached to B . The string passes over a small smooth fixed pulley at C , where C is vertically above A and \(\mathrm { AC } = 6 a\). A particle of mass \(\lambda m\), where \(\lambda\) is a positive constant, is attached to the other end of the string and hangs freely, vertically below C . The rod makes an angle \(\theta\) with the upward vertical, where \(0 \leqslant \theta \leqslant \pi\). You may assume that the particle does not interfere with the rod AB or the section of the string BC . \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system relative to a situation in which the rod AB is horizontal, and hence show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 2 m g a \sin \theta \left( \frac { 3 \lambda } { \sqrt { 10 - 6 \cos \theta } } - 1 \right) .$$
2. Show that \(\theta = 0\) and \(\theta = \pi\) are the only values of \(\theta\) for which the system is in equilibrium whatever the value of \(\lambda\).
3. Show that, if there is a third value of \(\theta\) for which the system is in equilibrium, then \(\frac { 2 } { 3 } < \lambda < \frac { 4 } { 3 }\).
4. Given that there are three positions of equilibrium, establish whether each of these positions is stable or unstable. It is given that, for small values of \(\theta\), $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } \approx 2 m g a \left[ \left( \frac { 3 } { 2 } \lambda - 1 \right) \theta - \left( \frac { 13 } { 16 } \lambda - \frac { 1 } { 6 } \right) \theta ^ { 3 } \right] .$$
5. Investigate the stability of the equilibrium position given by \(\theta = 0\) in the case when \(\lambda = \frac { 2 } { 3 }\).

1 A small object of mass 5 kg is attached to one end of each of two identical parallel light elastic strings. The upper ends of both strings are attached to a horizontal ceiling.
The object hangs in equilibrium at R , with the extension of each string being 0.1 m , as shown in Fig. 1. \begin{figure}[h] \end{figure}

1. Find the stiffness of each string. One of the strings is now removed and the object initially falls downwards. The object does not return to R at any point in the subsequent motion.
2. Suggest a reason why the object does not return to \(R\).

2 A light elastic string has natural length \(a\) and modulus of elasticity \(k m g\), where \(k > 2\). One end of the string is attached to a fixed point O . A particle P of mass \(m\) is attached to the other end of the string. P is held at rest a distance \(\frac { 3 } { 2 } a\) vertically below O . At time \(t\) after P is released, its vertical distance below O is \(y\).

1. Show that, while the string is in tension, the equation of motion of P is given by the differential equation \(\frac { d ^ { 2 } y } { d t ^ { 2 } } = ( k + 1 ) g - \frac { k g } { a } y\). A student transforms the differential equation in part (a) into the standard SHM equation \(\frac { d ^ { 2 } x } { d t ^ { 2 } } = - \omega ^ { 2 } x\).
2. - Find an expression for \(x\) in terms of \(y , k\) and \(a\).

6 Two identical light elastic strings, each of length \(l\) and modulus of elasticity \(\lambda m g\) are attached to a particle \(P\) of mass \(m\). The other end of the first string is attached to a fixed point A , and the other end of the second string is attached to a fixed point B . The points A and B are such that A is above and to the right of B and both strings are taut. The string attached to A makes an angle of \(30 ^ { \circ }\) with the vertical, and the string attached to B makes an angle of \(\theta ^ { \circ }\) with the horizontal, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{feb9a438-26b0-41d3-b044-6acd6efccde0-6_546_533_699_242} The system is in equilibrium in a vertical plane. The extension of the string attached to A is 0.9 l and the extension of the string attached to B is \(0.5 l\).

1. Explain how you know that APB is not a straight line.
2. Show that the elastic potential energy stored in string AP is \(k m g l\), where the value of \(k\) is to be determined correct to \(\mathbf { 3 }\) significant figures.

1 The end O of a light elastic string OA is attached to a fixed point.
Fiona attaches a mass of 1 kg to the string at A . The system hangs vertically in equilibrium and the length of the stretched string is 70 cm . Fiona removes the 1 kg mass and attaches a mass of 2 kg to the string at A . The system hangs vertically in equilibrium and the length of the stretched string is now 80 cm . Fiona then removes the 2 kg mass and attaches a mass of 5 kg to the string at A . The system hangs vertically in equilibrium.

1. Use the information given in the question to determine expected values for
   Fiona discovers that, when the mass of 5 kg is attached to the string at A , the length of the stretched string is greater than the expected length.
2. Suggest a reason why this has happened.

3 A young woman wishes to make a bungee jump. One end of an elastic rope is attached to her safety harness. The other end is attached to the bridge from which she will jump. She calculates that the stretched length of the rope at the bottom of her motion should be 20 m , she knows that her weight is 576 N and the stiffness of the elastic rope is \(90 \mathrm { Nm } ^ { - 1 }\). She has to calculate the unstretched length of rope required to perform the jump safely. She models the situation by assuming the following.

• The rope is of negligible mass.
• Air resistance may be neglected.
• She is a particle.
• She moves vertically downwards from rest.
• Her starting point is level with the fixed end of the rope.
• The length she calculates for the rope does not include any extra for attaching the ends.
  1. (A) Show that the greatest extension of the rope, \(X\), satisfies the equation \(X ^ { 2 } = 256\).
     (B) Hence determine the natural length of rope she needs.
  2. To remain safe she wishes to be sure that, if air resistance is taken into account, the stretched length of the rope of natural length determined in part (i) will not be more than 20 m . Advise her on this point.

3 One end of a light elastic spring of natural length 0.3 m is attached to a fixed point. A mass of 4 kg is attached to the other end of the spring. When the spring hangs vertically in equilibrium the extension of the spring is 0.02 m .

1. Determine the modulus of elasticity of the spring. A student calculates that if the mass of 4 kg is removed and replaced with a mass of 20 kg the extension of the spring will be 0.1 m .
2. Suggest a reason why this extension may not be 0.1 m .

12 A particle P of mass \(m\) is fixed to one end of a light elastic string of natural length \(l\) and modulus of elasticity 12 mg . The other end of the string is attached to a fixed point O . Particle P is held next to O and then released from rest.

1. Show that P next comes instantaneously to rest when the length of the string is \(\frac { 3 } { 2 } l\). The string first becomes taut at time \(t = 0\). At time \(t \geqslant 0\), the length of the string is \(l + x\), where \(x\) is the extension in the string.
2. Show that when the string is taut, \(x\) satisfies the differential equation $$\ddot { \mathrm { x } } + \omega ^ { 2 } \mathrm { x } = \mathrm { g } \text {, where } \omega ^ { 2 } = \frac { 12 \mathrm {~g} } { \mathrm { I } } \text {. }$$
3. By using the substitution \(x = y + \frac { g } { \omega ^ { 2 } }\), solve the differential equation to show that the time when the string first becomes slack satisfies the equation $$\cos \omega \mathrm { t } - \sqrt { \mathrm { k } } \sin \omega \mathrm { t } = 1$$ where \(k\) is an integer to be determined.

1. The diagram shows a spring of natural length 0.15 m enclosed in a smooth horizontal tube. One end of the spring \(A\) is fixed and the other end \(B\) is compressed against a ball of mass \(0 \cdot 1 \mathrm {~kg}\). \includegraphics[max width=\textwidth, alt={}, center]{b430aa50-27e3-46f7-afef-7b8e75d46e1f-2_241_714_639_632}

Initially, the ball is held in equilibrium by a force of 21 N so that the compressed length of the spring is \(\frac { 2 } { 5 }\) of its natural length.

1. Calculate the modulus of elasticity of the spring.
2. The ball is released by removing the force. Determine the speed of the ball when the spring reaches its natural length. Give your answer correct to two significant figures.

1. The diagram below shows a light spring of natural length 1.2 m and modulus of elasticity 84 N . One end of the spring \(A\) is fixed and the other end is attached to an object \(P\) of mass 4 kg . \includegraphics[max width=\textwidth, alt={}, center]{ae23a093-1419-4be4-8285-951650dc5a35-06_542_451_466_808}

Initially, \(P\) is held at rest with the spring stretched to a total length of 2.2 m and \(A P\) vertical.

1. Show that the elastic energy stored in the spring is 35 J .
2. The object \(P\) is then released. Find the speed of \(P\) at the instant when the elastic energy in the spring is reduced to \(5 \cdot 6 \mathrm {~J}\).

6. The diagram shows a particle \(P\), of mass 4 kg , lying on a smooth horizontal surface. It is attached by two light springs to fixed points \(A\) and \(B\), where \(A B = 2.8 \mathrm {~m}\).
Spring \(A P\) has natural length 0.8 m and modulus of elasticity 60 N .
Spring \(P B\) has natural length 1.2 m and modulus of elasticity 30 N . \includegraphics[max width=\textwidth, alt={}, center]{b9c63cb4-d446-4548-be42-e30b10cb4b99-5_231_1253_612_404} When \(P\) is in equilibrium, it is at the point \(C\).

1. Show that \(A C = 1 \mathrm {~m}\).
2. The particle \(P\) is pulled horizontally and is initially held at rest at the midpoint of \(A B\). The system is then released.
   1. Show that \(P\) performs Simple Harmonic Motion about centre \(C\) and find the period of its motion.
   2. Determine the shortest time taken for \(P\) to reach a position where there is no tension in the spring \(A P\). \section*{END OF PAPER}

6. The diagram on the left shows a train of mass 50 tonnes approaching a buffer at the end of a straight horizontal railway track. The buffer is designed to prevent the train from running off the end of the track. The buffer may be modelled as a light horizontal spring \(A B\), as shown in the diagram on the right, which is fixed at the end \(A\). The train strikes the buffer so that \(P\) makes contact with \(B\) at \(t = 0\) seconds. While \(P\) is in contact with \(B\), an additional resistive force of \(250000 v \mathrm {~N}\) will oppose the motion of the train, where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of the train at time \(t\) seconds. The spring has natural length 1 m and modulus of elasticity 312500 N . At time \(t\) seconds, the compression of the spring is \(x\) metres. \includegraphics[max width=\textwidth, alt={}, center]{d7f600c5-af4a-4708-bfd9-92b37a95c634-7_358_1506_824_283}

1. Show that, while \(P\) is in contact with \(B\), \(x\) satisfies the differential equation $$4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 20 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 25 x = 0$$
2. Given that, when \(P\) first makes contact with \(B\), the speed of the train is \(U \mathrm {~ms} ^ { - 1 }\), find an expression for \(x\) in terms of \(U\) and \(t\).
3. When the train comes to rest, the compression of the buffer is 0.3 m . Determine the speed of the train when it strikes the buffer.
4. State which type of damping is described by the motion of \(P\). Give a reason for your answer.

6. The diagram shows a playground ride consisting of a seat \(P\), of mass 12 kg , attached to a vertical spring, which is fixed to a horizontal board. When the ride is at rest with nobody on it, the compression of the spring is 0.05 m . \includegraphics[max width=\textwidth, alt={}, center]{3efc4ef6-8a80-4267-8e95-733200e875c5-4_305_654_1032_667} The spring is of natural length 0.75 m and modulus of elasticity \(\lambda\).

1. Find the value of \(\lambda\). The seat \(P\) is now pushed vertically downwards a further 0.05 m and is then released from rest.
2. Show that \(P\) makes Simple Harmonic oscillations of period \(\frac { \pi } { 7 }\) and write down the amplitude of the motion.
3. Find the maximum speed of \(P\).
4. Calculate the speed of \(P\) when it is at a distance 0.03 m from the equilibrium position.
5. Find the distance of \(P\) from the equilibrium position 1.6 s after it is released.[3]
6. State one modelling assumption you have made about the seat and one modelling assumption you have made about the spring.

9 In this question use \(g = 9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
A light elastic string has one end attached to a fixed point, \(A\), on a rough plane inclined at \(30 ^ { \circ }\) to the horizontal. The other end of the string is attached to a particle, \(P\), of mass 2 kg .
The elastic string has natural length 1.3 metres and modulus of elasticity 65 N .
The particle is pulled down the plane in the direction of the line of greatest slope through \(A\).
The particle is released from rest when it is 2 metres from \(A\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{4fdb2637-6368-422c-99da-85b80efe31c5-14_549_744_861_785} The coefficient of friction between the particle and the plane is 0.6
After the particle is released it moves up the plane.
The particle comes to rest at a point \(B\), which is a distance, \(d\) metres, from \(A\). 9

1. Show that the value of \(d\) is 1.4.
   [0pt] [7 marks] 9
2. Determine what happens after \(P\) reaches the point \(B\). Fully justify your answer.
   [0pt] [3 marks]

1. A spring of natural length \(a\) has one end attached to a fixed point \(A\). The other end of the spring is attached to a package \(P\) of mass \(m\).
   The package \(P\) is held at rest at the point \(B\), which is vertically below \(A\) such that \(A B = 3 a\).
   After being released from rest at \(B\), the package \(P\) first comes to instantaneous rest at \(A\). Air resistance is modelled as being negligible.

By modelling the spring as being light and modelling \(P\) as a particle,

1. show that the modulus of elasticity of the spring is \(2 m g\)
2. 1. Show that \(P\) attains its maximum speed when the extension of the spring is \(\frac { 1 } { 2 } a\)
   2. Use the principle of conservation of mechanical energy to find the maximum speed, giving your answer in terms of \(a\) and \(g\). In reality, the spring is not light.
3. State one way in which this would affect your energy equation in part (b).

1. A light elastic string has natural length \(2 a\) and modulus of elasticity \(2 m g\). One end of the string is attached to a fixed point \(A\) on a horizontal ceiling. The other end is attached to a particle \(P\) of mass \(m\).

The particle \(P\) hangs in equilibrium at the point \(E\), where \(A E = 3 a\).
The particle \(P\) is then projected vertically downwards from \(E\) with speed \(\frac { 3 } { 2 } \sqrt { a g }\) Air resistance is assumed to be negligible.
Find the elastic energy stored in the string, when \(P\) first comes to instantaneous rest. Give your answer in the form kmga, where \(k\) is a constant to be found.

1. A particle \(P\) of mass \(m\) is attached to one end of a light elastic string of natural length \(a\) and modulus of elasticity \(3 m g\).

The other end of the string is attached to a fixed point \(O\) on a ceiling.
The particle hangs freely in equilibrium at a distance \(d\) vertically below \(O\).

1. Show that \(d = \frac { 4 } { 3 } a\). The point \(A\) is vertically below \(O\) such that \(O A = 2 a\).
   The particle is held at rest at \(A\), then released and first comes to instantaneous rest at the point \(B\).
2. Find, in terms of \(g\), the acceleration of \(P\) immediately after it is released from rest.
3. Find, in terms of \(g\) and \(a\), the maximum speed attained by \(P\) as it moves from \(A\) to \(B\).
4. Find, in terms of \(a\), the distance \(O B\).

1. The points \(A\) and \(B\) lie on a smooth horizontal surface with \(A B = 4.5 \mathrm {~m}\).

A light elastic string has natural length 1.5 m and modulus of elasticity 15 N . One end of the string is attached to \(A\) and the other end of the string is attached to \(B\). A particle, \(P\), of mass 0.2 kg , is attached to the stretched string so that \(A P B\) is a straight line and \(A P = 1.5 \mathrm {~m}\). The particle rests in equilibrium on the surface. The particle is now moved directly towards \(A\) and is held on the surface so \(A P B\) is a straight line with \(A P = 1 \mathrm {~m}\). The particle is released from rest.

1. Prove that \(P\) moves with simple harmonic motion.
2. Find
   1. the maximum speed of \(P\) during the motion,
   2. the maximum acceleration of \(P\) during the motion.
3. Find the total time, in each complete oscillation of \(P\), for which the speed of \(P\) is greater than \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).